L2 error estimate under a Dirichlet Poisson smoothing index

Put $e:=u-u_h$. If $e=0$ in $L^2$, the conclusion holds. Otherwise apply [F1] with $g=e$ to obtain $\zeta \in H_0^1(\mathrm{\Omega })\cap H^{1+s}(\mathrm{\Omega })$ with $\mathrm{\parallel }\zeta {\mathrm{\parallel }}_{H^{1+s}}\le C_{\mathrm{r}\mathrm{e}\mathrm{g}}\mathrm{\parallel }e{\mathrm{\parallel }}_2$. Testing its weak equation on $C_c^{\mathrm{\infty }}(\mathrm{\Omega })\subset H_0^1(\mathrm{\Omega })$ gives $-\mathrm{\Delta }\zeta =e$ distributionally.

Let $I_{h}u$ be the global interpolant and put $w_h:=I_{h}u-u_h\in V_h^{CR,0}$. The jump of $e$ decomposes as $[e]=[u-I_{h}u]+[w_h]$. The local face estimate in [F3], using the common trace of $u$ on interior facets and its zero boundary trace, gives ${\sum }_F{h}_F^{-1}\mathrm{\parallel }[u-I_{h}u{]}_F{\mathrm{\parallel }}_2^2\le C{h}^{2r}\mathrm{\mid }u{\mathrm{\mid }}_{H^{1+r}}^2\mathrm{.}$ By [F4] and the triangle inequality (which follows for the finite direct sum of face $L^2$ spaces by expanding squares and applying [F8]), ${({\sum }_F{h}_F^{-1}\mathrm{\parallel }[w_h{]}_F{\mathrm{\parallel }}_2^2)}^{1\mathrm{/}2}\le C\mathrm{\parallel }{\mathrm{\nabla }}_h{w}_h{\mathrm{\parallel }}_2\le C(\mathrm{\parallel }{\mathrm{\nabla }}_h(I_{h}u-u){\mathrm{\parallel }}_2+\mathrm{\parallel }{\mathrm{\nabla }}_{h}e{\mathrm{\parallel }}_2)\le C{h}^r\mathrm{\mid }u{\mathrm{\mid }}_{H^{1+r}},$ where [F3] bounds the first term and [F5] the second. Combining these estimates yields ${({\sum }_F{h}_F^{-1}\mathrm{\parallel }[e{]}_F{\mathrm{\parallel }}_2^2)}^{1\mathrm{/}2}\le C{h}^r\mathrm{\mid }u{\mathrm{\mid }}_{H^{1+r}}\mathrm{.}$ Each $[e{]}_F$ has zero face mean: on interior facets the means of $u$ and $u_h$ match, and on boundary facets both are zero.

Apply [F2] on every cell with $w=e{\mathrm{\mid }}_K$ and sum. Write $[e{]}_F$ for the difference of its two traces on an interior facet and its trace on a boundary facet. Since $-\mathrm{\Delta }\zeta =e$, $\mathrm{\parallel }e{\mathrm{\parallel }}_2^2={\sum }_K{\int }_K\mathrm{\nabla }e\cdot \mathrm{\nabla }\zeta dx-{\sum }_F{\int }_F({\gamma }_F\mathrm{\nabla }\zeta \cdot n_F)[e{]}_{F}dS\mathrm{.}$ For interior $F$, $n_F$ is the outward normal of either fixed plus cell; for a boundary facet it is the outward cell normal. The global traces of $\mathrm{\nabla }\zeta$ match across interior facets by [F2].

By the trace fluctuation bound [F2], Holder [F8] on the disjoint union of facets gives $\mid {\sum }_F{\int }_F({\gamma }_F\mathrm{\nabla }\zeta \cdot n_F)[e{]}_{F}dS\mid \le {({\sum }_F{h}_F\mathrm{\parallel }{q}_F-{\mathrm{avg}}_F{q}_F{\mathrm{\parallel }}_2^2)}^{1\mathrm{/}2}{({\sum }_F{h}_F^{-1}\mathrm{\parallel }[e{]}_F{\mathrm{\parallel }}_2^2)}^{1\mathrm{/}2}\le C{h}^s\mathrm{\mid }\zeta {\mathrm{\mid }}_{H^{1+s}}{h}^r\mathrm{\mid }u{\mathrm{\mid }}_{H^{1+r}},$ where $q_F$ is the corresponding normal trace. The last inequality uses that each cell has $d+1$ facets, $h_F\le h$, and the cell Gagliardo seminorms sum to at most the global seminorm.

Split the volume term in step 2.1 as ${\sum }_K{\int }_K\mathrm{\nabla }e\cdot \mathrm{\nabla }(\zeta -I_h\zeta )+{\tilde{a}}_h(e,I_h\zeta )\mathrm{.}$ By [F7], linearity, and the discrete equation $a_h(u_h,I_h\zeta )=F_h(I_h\zeta )$, the second term is ${\tilde{a}}_h(u,I_h\zeta )-F_h(I_h\zeta )$, the consistency residual in [F6]. Holder [F8] on the cells bounds the first term, and [F3] applied with exponent $s$ together with [F5] gives $\mathrm{\parallel }{\mathrm{\nabla }}_{h}e{\mathrm{\parallel }}_2\mathrm{\parallel }{\mathrm{\nabla }}_h(\zeta -I_h\zeta ){\mathrm{\parallel }}_2\le C{h}^r\mathrm{\mid }u{\mathrm{\mid }}_{H^{1+r}}{h}^s\mathrm{\mid }\zeta {\mathrm{\mid }}_{H^{1+s}}\mathrm{.}$ Together with [F6], its magnitude is at most $C{h}^{r+s}\mathrm{\mid }u{\mathrm{\mid }}_{H^{1+r}}\mathrm{\mid }\zeta {\mathrm{\mid }}_{H^{1+s}}$.

Substitute the bounds from steps 2.2 and 3.1 in the identity of step 2.1. This gives $\mathrm{\parallel }e{\mathrm{\parallel }}_2^2\le C{h}^{r+s}\mathrm{\mid }u{\mathrm{\mid }}_{H^{1+r}}\mathrm{\mid }\zeta {\mathrm{\mid }}_{H^{1+s}}\le C{C}_{\mathrm{r}\mathrm{e}\mathrm{g}}{h}^{r+s}\mathrm{\mid }u{\mathrm{\mid }}_{H^{1+r}}\mathrm{\parallel }e{\mathrm{\parallel }}_2$ by step 1.1. Since $e\mathrm{\ne }0$, divide by $\mathrm{\parallel }e{\mathrm{\parallel }}_2$ to obtain the stated result.